$A$ can filled with water is revolved in a vertical circle of radius $r$ with constant speed such that the water just does not fall out. The time period of revolution is ($g=$ acceleration due to gravity).

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a horizontal velocity $v_{o}$ at the lowest point $A$ such that it completes a circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point,$C$. This is shown in the figure. Obtain an expression for: $(i) v_{o}$; $(ii)$ the speeds at points $B$ and $C$; $(iii)$ the ratio of the kinetic energies $(K_{B} / K_{C})$ at $B$ and $C$. Comment on the nature of the trajectory of the bob after it reaches the point $C$.

$A$ weightless string can support a tension up to $30 \,N$. $A$ stone of mass $0.5 \,kg$ is tied to its one end and is revolved in a circular path of radius $2 \,m$ in a vertical plane. Then the maximum angular velocity of the stone will be (acceleration due to gravity $g=10 \,m/s^2$)

Length of a simple pendulum is $1 \,m$. When its bob is at its lowest point its velocity is $7 \,ms^{-1}$. If the bob leaves its circular path at a height $h$ above the centre of the circle, then the value of $h$ is $(g=10 \,ms^{-2})$. (in $\,m$)

$A$ bob of mass $m$ suspended by a light string of length $L$ is whirled in a vertical circle as shown in the figure. What will be the trajectory of the particle if the string is cut at $(a)$ point $B$,$(b)$ point $C$,and $(c)$ point $X$?

The string of a pendulum of length $l$ is displaced through $90^{\circ}$ from the vertical and released. The minimum strength of the string required to withstand the tension as the pendulum passes through the mean position is: